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Multiplying both sides of the last equation by 2  dt, and integrating we
MY= -\dt)
But  = 0 when 6 = 0, therefore c = ^.   Hence d                                                    a
a>2 = ^(1-0080). a
DISCUSSION.  It will be observed that this result is already given by the energy equation.
When 6 = ^, co2 =  , therefore the door strikes the floor with an 2i                 d
angular velocity of t /_ .   Thus the greater a the less the angular velocity
V   a
with which the door strikes the floor. On the other hand the linear velocity with which the end of the door strikes the floor increases with a, since
= 3ag, whentf = ^-
1.   Discuss the motion of the falling bodies in Atwood's machine, supposing the pulley to rotate without slipping.
2.   In the problem discussed in the first illustrated example take into account the resistance of the air, supposing the resistance to be proportional to the angular velocity of the wheel.
3.   A flywheel which is making 400 revolutions per minute and which is subject to a constant torque of 50 pounds-foot comes to rest after making 1500 revolutions.    Find the moment of inertia of the wheel, the angular acceleration and the time taken in coming to rest.   The angular acceleration is supposed to be constant.
4.   A flywheel which is subject to a constant torque of 5000 dynes-centimeter starts from rest and makes 2000 revolutions in 4 minutes, Find the angular acceleration and the moment of inertia.